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3 years ago

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126 divide 6' long division

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swat323 years ago

3 0

Http://www.free-hosting.biz/division/126-divided-6.html

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Which algebraic expression is equivalent to the expression below?

jeyben [28]

Answer:

<h2>A. 10x + 30</h2>

Step-by-step explanation:

2(3 + 5x) + 24 = 6 + 10x + 24 = 10x +(6 + 24) = 10x + 30

4 0

3 years ago

Read 2 more answers

Minnie has 4 different stamps. She needs 3 to mail a large letter. How many ways can she pick the 3 that will be used?

timurjin [86]

Number of ways of picking a smaller number out of a bigger number is called combination.
Number of way of picking 3 stamps out of 4 is 4C3 (4 combination 3) = 4! / (3! x (4 - 3)!) = 4! / (3! x 1!) = (4 x 3 x 2 x 1) / (3 x 2 x 1 x 1) = 24 / 6 = 4 ways.

7 0

3 years ago

An alloy is composed of nickel, zinc, and copper in a ration of 7:2:9. How many kilograms of each metal are needed to make 3.6kg

motikmotik

Well, the short answer is, we divide that 3.6 by (7+2+9) and then we give as many pieces at the ratio to each, so let's do so,

$\bf \stackrel{nickel}{7}~~:~~\stackrel{zinc}{2}~~:~~\stackrel{copper}{9}
\\\\\\
\stackrel{nickel}{7\cdot \frac{3.6}{7+2+9}}~~:~~\stackrel{zinc}{2\cdot \frac{3.6}{7+2+9}}~~:~~\stackrel{copper}{9\cdot \frac{3.6}{7+2+9}}
\\\\\\
\stackrel{nickel}{7\cdot 0.2}~~:~~\stackrel{zinc}{2\cdot 0.2}~~:~~\stackrel{copper}{9\cdot 0.2}
\\\\\\
\stackrel{nickel}{1.4}~~:~~\stackrel{zinc}{0.4}~~:~~\stackrel{copper}{1.8}$

6 0

3 years ago

Read 2 more answers

Alma is estimating the proportion of students in her school district who, in the past month, read at least 1 book. From a random

Maurinko [17]

Answer:

D)

$(0.64 +1.645 \sqrt{\frac{0.64(0.36)}{50} })$

90 percent confidence interval for the proportion of all students in her district who read at least 1 book last month

(0.52846 , 0.75166)

Step-by-step explanation:

<u><em>Step:-1</em></u>

Given that the random sample size 'n' = 50

The sample proportion

$p^{-} = \frac{32}{50} = 0.64$

90 percent confidence interval for the proportion of all students in her district who read at least 1 book last month

$(p^{-} - Z_{0.90} \sqrt{\frac{p(1-p)}{n}, } , p^{-} +Z_{0.90} \sqrt{\frac{p(1-p)}{n} })$

<u><em>Step(ii):-</em></u>

Level of significance = 0.90

Z₀.₉₀ = 1.645

$(0.64 - 1.645 \sqrt{\frac{0.64(1-0.64)}{50}, } , 0.64 +1.645 \sqrt{\frac{0.64(1-0.64)}{50} })$

$(0.64 - 1.645 \sqrt{\frac{0.64(0.36)}{50}, } , 0.64 +1.645 \sqrt{\frac{0.64(0.36)}{50} })$

(0.64 -1.645(0.06788) , (0.64 + 1.645(0.06788)

(0.52846 , 0.75166)

<u><em>Final answer:-</em></u>

90 percent confidence interval for the proportion of all students in her district who read at least 1 book last month

(0.52846 , 0.75166)

5 0

3 years ago

12÷478<br> find two estimates <br> plz help thank you<br> ill give you 12 points

BaLLatris [955]

10/480 equals about .02083 and 12/478 equals about <span>.02510. </span>

6 0

3 years ago